# Question 7 2015 STEP 1

Please can someone explain why it isn’t a^3/9 for all a greater than or equal to two. In the mark scheme it says it’s always bigger than 3a-6 but then let’s M(a) be 3a-6 ? Mark scheme at: https://correspondence.maths.org/sites/correspondence.maths.org/files/fe...

### 2015 S1 Q7

If $\frac a 3>1$ then the local maximum doesn't lie in the range $-\frac 1 3 \le x \le 1$. The maximal value will be at one of the end points.

So if $\frac a 3 > 1 \implies a>3$ then the maximum value will be at $f(1)=3a-6$. $\frac{a^3}9$ will be a bigger value but $x=\frac a 3$ will not lie in the range of $x$ when $a>3$.